Mathematical Solid Geometry Concluding Statements

Cylinder: the radius is r, the height is 2h, and the radius of the receiving ball r=(r square + h squared) square, for example, r=3, h=5

Then r=5 cone: cone top angle 2, bus 2L, external ball radius r=l cos

Cuboid: length 2a, width 2b, height 2c, radius of the outside ball r = (a square + b square + c squared) open.

Triangular pyramid: triangular pyramid ABCD, with the triangle ABC as the bottom surface and D as the vertex (ABCD four points are based on the coordinate table.

show). 1. Find the outer center point O of δabc (i.e., the intersection point of the perpendicular line on any two sides);

Second, the perpendicular line of the ABC of the surface O is crossed;

3. Set a point E on the perpendicular line, so that ED=EA (or ED=EB, or ED=EC) can list an equation and find the coordinates of the point E;

Outside ball radius r=ea=eb=ec=ed

Multiple (four or more) pyramids: The base polygon must be regular polygonal (quadrilaterals can be square or rectangular.)

shape), 1. Find the outer center of the bottom polygon o (i.e., the intersection of the perpendicular lines on any two sides), if found.

If there is no external center, then this polyangular pyramid has no external ball;

Second, the point o to do the perpendicular line of the bottom surface;

3. Set a point E on the perpendicular line so that ed=ea (or ed=eb, or ed=ec...).can be listed.

to find the coordinates of the point e;

Outside ball radius r=ea=eb=ec=ed

Prism: 1. Find the outer center of the two bottom surfaces of the prism;

2. Connect the outer center of the two bottom surfaces and find the midpoint m of the line;

3. The line connecting the point m with any vertex of the prism is the radius r of the outside ball.

External. Cylinder: The square of the diameter of the cylinder above the ground and the square of the square divided by 2 are too complicated, you just need to find the common plane of the geometry in the focal point of the outer ball and convert it into a plane figure.

The inner column of the garden has to be divided into situations, and it doesn't seem to be much of a test.

The triangular pyramid is even more difficult to say, and you didn't say the positive pyramid, so you can't express it.

Axiom 1: If two points on a straight line are in a plane, then all the points in the line are in that plane.

Axiom 2: If two planes have a common point, then they have other common points, and the set of all these common points is a straight line passing through this common point.

Axiom 3: After passing three points that are not on the same straight line, there is and only one plane.

Corollary 1: After a straight line and a point outside of this straight line, there is one and only one plane.

Corollary 2: After two intersecting straight lines, there is one and only one plane.

Corollary 3: After two parallel straight lines, there is only one plane.

Are you going to carry it? Then it doesn't work. You can't use it flexibly if you memorize it.

The key to solving the problem of the external ball of the geometry is to find the center of the sphere, and then the problem is solved naturally, of course, the geometry at this time is mostly geometry with very good symmetry, such as regular triangular pyramid, regular tetrahedron, cube, cuboid, etc. You need to understand the characteristics of these regular geometries! On the other hand, the inscribed sphere of the geometry should grasp the point that the distance from the center of the sphere to each surface is the radius of the ball, and the volume method is usually used to solve the problem.

You can sort these out without delaying time, which is much better than when you memorize them.

n pyramid, n prismatic volume.

v=1 3*base area*height.

The cube is connected to the outside (edge length is a).

Inner ball volume v=4 3* *a 2) 3

Outer ball volume v=4 3* *1 2) 3 The derivation process is a derivative, and then converted into a proportional sequence to find the limit, it is recommended that the landlord try the derivation process when he is free, I remember that I used the derivative to derive these formulas by myself when I was bored in high school, and now I have put down mathematics for almost 3 years, and I am still impressed. Good luck with the college entrance examination!

There are two keys to learning solid geometry well:

1. Graphics: It is very important to learn not only to read pictures, but also to learn to draw, and to cultivate their spatial imagination ability through reading and drawing.

At the beginning, you need to look at and think about the model, for example: your classroom is a cuboid, the pyramid is a pyramid, the pen is a straight line, the desktop is a flat surface, think about the straight line inside.

In the medium term, these models come to mind.

In the later stage, I basically figured out the figure, and drew the figure.

2. Language: Many students can think about the problem clearly, but when it falls on paper, they can't speak. A word to remember:

Geometric language is the most important thing to say with evidence and reason. In other words, don't say anything that doesn't have a basis, and don't say anything that doesn't conform to the theorem.

As for how to prove the problem of solid geometry, we can study it from the following two perspectives:

1) Classify all theorems in geometry: the classification according to the known conditions of the theorem is the property theorem, and the classification according to the conclusion of the theorem is the decision theorem.

For example, if two straight lines parallel to the same straight line are parallel, it can be regarded as either the property theorem of the parallel nature of two straight lines, or it can be regarded as it.

Cheng is a decision theorem in which two straight lines are parallel.

For example, if two planes are parallel and intersect the third plane at the same time, then their intersection lines are parallel. It is both a theorem of the nature of two planes that are parallel.

Again, two judgment theorems with parallel straight lines. In this way, we can find what we need, for example: we want to prove a straight line.

and perpendicular to the plane, the following theorem can be used:

1) The determination theorem that lines and planes are perpendicular.

2) Two parallel perpendicular to the same plane.

3) A straight line and two parallel planes are perpendicular at the same time.

2) Be clear about what you want to do

Be sure to know what you're going to do! Before proofing, you must design a good route, clarify the purpose of each step, learn to make bold assumptions, and reason carefully.

3. The space vector method can avoid tedious logical reasoning, and you can learn it.

4. The college entrance examination is almost the same for several exams, generally 1-2 small questions and 1 big question.

Depending on what province you are, you can see what type of three-dimensional geometry is generally tested in the college entrance examination over the years, and you can take your time with poor imagination, because my previous imagination was not very good.

Mathematics stereoscopic geometry in the college entrance examination is a big question, and it doesn't matter if you can't imagine it. High school mathematics has already thought of a way for you to --- vector method, using vector method without geometric imagination, but the steps are more cumbersome and ordinary people are unwilling to use, but it is still more reliable and practical. Come on!

Everything will be fine.

Quick solution: start with the proof question, do 5 to 10 questions for each theorem question type in the order in which the theorem appears, draw another picture for each question, do not choose the difficult one, do it, memorize the theorem and then use it well, and other question types are self-defeating.

A lot of students bai

The subconscious mind will make du reasoning like this::

1) My geometry of the body is not good - >

2) Because I don't have a good spatial imagination.

Force->3) Good spatial imagination should be innate->

4) Therefore, I am not good at stereo geometry because I am naturally "stupider" than others in this regard ->5) Therefore, no matter how hard I try, it is in vain.

And many teachers can't teach the law, so that those children who have worked hard still can't make progress, so they believe in the above reasoning even more, and eventually it becomes a vicious circle.

In fact, as long as you master the right method, you can improve the problem-solving ability of three-dimensional geometry by using Li Zeyu's three tricks of translation-specialization-staring at the target.

If you can't see clearly, you can ask again.

Solid geometry, as a key and hot spot in mathematics in the college entrance examination over the years, has always been one of the compulsory contents. Looking at the mathematics papers of the college entrance examination across the country, we can intuitively see that there must be a solution question or several fill-in-the-blank questions and multiple-choice questions related to three-dimensional geometry every year.

Although many teachers often emphasize the importance of solid geometry, many candidates do not score well on this piece of content. At the same time, this also reminds candidates who are about to take the college entrance examination that if they can thoroughly understand the knowledge content of three-dimensional geometry, they will definitely improve their mathematics scores in the college entrance examination.

In the mathematics of the college entrance examination, the objective questions related to three-dimensional geometry mainly examine the determination of the basic position relationship, as well as the calculation of the angle, distance and volume of the column, cone and ball, which are short, concise, novel and unique, unique in design, and high in ability and intention. The solution is mainly to prove the position relationship of the spatial line and surface and the calculation of the relevant quantity relationship, such as the determination and proof of the parallel and perpendicular of the spatial line and surface, and the calculation of the angle and distance of the line and surface.

Through the setting of questions related to three-dimensional geometry, the candidates' spatial imagination ability, reasoning and argumentation ability, as well as naturalization and transformation ability can be well examined, reflecting the function of mathematics in the college entrance examination to select talents.

College Entrance Examination Mathematics Solid Geometry, Typical Example Problem Analysis 1:

As shown in the figure, in the quadrangular pyramid P ABCD, the bottom surface is a right-angled trapezoidal ABCD, where AD AB, CD AB, AB 4, CD 2, and the side PAD is an equilateral triangle with a side length of 2, and it is perpendicular to the bottom ABCD, and E is the midpoint of PA

1) Verification: DE planar PBC;

2) Find the volume of triangular pyramid A PBC

In the right-angled trapezoidal ABCD, CD AB, and AB 4, CD 2, so BF is c

So the quadrilateral bcdf is a parallelogram

So df bc

In PAB, PE EA, AF FB, so EF PB

And because df ef f, pb bc b, then the plane def plane pbc

Because of the de plane def, so the de plane pbc

2) Take the midpoint O of AD and connect Po

In pad, pa pd ad 2, so po ad, po

And because the plane pad plane abcd, the plane pad plane abcd ad, so po plane abcd

For the three-dimensional geometry problems related to parallel and perpendicular, we must dig deep into the conditions of the problem, combine the relevant property theorem, use the properties of the problem conditions to add auxiliary lines (or surfaces) appropriately, and connect the conclusions to gradually find the solution ideas.

It is worth noting that the three-perpendicular line theorem and its inverse theorem are used most frequently in college entrance examination questions, and should be given priority when proving that the line is perpendicular.

This is really my favorite question, if 20 years ago I would have given the answer soon, but now it doesn't, I have forgotten the mutual conditions, I have to pick it up again and study, I will re-read the books, consolidate the knowledge, replenish the energy, and come back to you in a few days to give you the answer, thank you for the question.

According to the meaning of the title and the first question, it is easy to prove that there are two lines perpendicular to the two faces of the ridge line in the plane AEF.

It's too simple, use the PA vertical ABCD to launch the PA vertical AE, and then prove the AD vertical AE, and then get the AE vertical PAD, and it's done.

PA 丄 ABCD (projection).

Therefore, ae 丄 pa

and ead=90° (self-proof).

Hence the ae 丄ad

i.e. ae pad

Again ae aef

Hence the aef pad

Seek to adopt, adopt, adopt.

What is the problem? Why didn't I see it.